For the system `3A + 2BhArr` C, the expression for equilibrium constant is

Options

A. `([3A]xx[2B])/([C])`
B. `([A]^3xx[B])/([C])`
C. `([C])/([A]^3xx[B]^2)`
D. `([C])/([3A]xx[2B])`

Answer

To derive the expression for the equilibrium constant for the reaction \( 3A + 2B \rightleftharpoons C \), we will follow these steps:

### Step 1: Write the balanced chemical equation.
The balanced chemical equation is given as:
\[ 3A + 2B \rightleftharpoons C \]

### Step 2: Identify the products and reactants.
In this reaction:
- **Reactants**: \( A \) and \( B \)
- **Product**: \( C \)

### Step 3: Write the general expression for the equilibrium constant \( K \).
The equilibrium constant \( K \) is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.

### Step 4: Apply the stoichiometric coefficients to the equilibrium expression.
For the reaction:
- The stoichiometric coefficient of \( C \) is 1.
- The stoichiometric coefficient of \( A \) is 3.
- The stoichiometric coefficient of \( B \) is 2.

Thus, the expression for the equilibrium constant \( K \) can be written as:
\[ K = \frac{[C]^1}{[A]^3[B]^2} \]

### Step 5: Simplify the expression.
Since the exponent for \( C \) is 1, we can simplify the expression to:
\[ K = \frac{[C]}{[A]^3[B]^2} \]

### Final Answer:
The expression for the equilibrium constant \( K \) for the reaction \( 3A + 2B \rightleftharpoons C \) is:
\[ K = \frac{[C]}{[A]^3[B]^2} \]

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